Integrand size = 29, antiderivative size = 75 \[ \int \frac {\left (d-e x^2\right )^{5/2}}{\left (d^2-e^2 x^4\right )^{3/2}} \, dx=\frac {2 x \sqrt {d-e x^2}}{\sqrt {d^2-e^2 x^4}}-\frac {\text {arctanh}\left (\frac {\sqrt {e} x \sqrt {d-e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{\sqrt {e}} \] Output:
2*x*(-e*x^2+d)^(1/2)/(-e^2*x^4+d^2)^(1/2)-arctanh(e^(1/2)*x*(-e*x^2+d)^(1/ 2)/(-e^2*x^4+d^2)^(1/2))/e^(1/2)
Time = 4.89 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.49 \[ \int \frac {\left (d-e x^2\right )^{5/2}}{\left (d^2-e^2 x^4\right )^{3/2}} \, dx=\frac {2 x \sqrt {d^2-e^2 x^4}}{\sqrt {d-e x^2} \left (d+e x^2\right )}+\frac {\log \left (-d+e x^2\right )}{\sqrt {e}}-\frac {\log \left (d e x-e^2 x^3+\sqrt {e} \sqrt {d-e x^2} \sqrt {d^2-e^2 x^4}\right )}{\sqrt {e}} \] Input:
Integrate[(d - e*x^2)^(5/2)/(d^2 - e^2*x^4)^(3/2),x]
Output:
(2*x*Sqrt[d^2 - e^2*x^4])/(Sqrt[d - e*x^2]*(d + e*x^2)) + Log[-d + e*x^2]/ Sqrt[e] - Log[d*e*x - e^2*x^3 + Sqrt[e]*Sqrt[d - e*x^2]*Sqrt[d^2 - e^2*x^4 ]]/Sqrt[e]
Time = 0.33 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1396, 298, 224, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (d-e x^2\right )^{5/2}}{\left (d^2-e^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1396 |
\(\displaystyle \frac {\sqrt {d-e x^2} \sqrt {d+e x^2} \int \frac {d-e x^2}{\left (e x^2+d\right )^{3/2}}dx}{\sqrt {d^2-e^2 x^4}}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\sqrt {d-e x^2} \sqrt {d+e x^2} \left (\frac {2 x}{\sqrt {d+e x^2}}-\int \frac {1}{\sqrt {e x^2+d}}dx\right )}{\sqrt {d^2-e^2 x^4}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {d-e x^2} \sqrt {d+e x^2} \left (\frac {2 x}{\sqrt {d+e x^2}}-\int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}\right )}{\sqrt {d^2-e^2 x^4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {d-e x^2} \sqrt {d+e x^2} \left (\frac {2 x}{\sqrt {d+e x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {e}}\right )}{\sqrt {d^2-e^2 x^4}}\) |
Input:
Int[(d - e*x^2)^(5/2)/(d^2 - e^2*x^4)^(3/2),x]
Output:
(Sqrt[d - e*x^2]*Sqrt[d + e*x^2]*((2*x)/Sqrt[d + e*x^2] - ArcTanh[(Sqrt[e] *x)/Sqrt[d + e*x^2]]/Sqrt[e]))/Sqrt[d^2 - e^2*x^4]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x _Symbol] :> Simp[(a + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a* e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {\left (-\ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right ) \sqrt {e \,x^{2}+d}+2 x \sqrt {e}\right ) \sqrt {-e^{2} x^{4}+d^{2}}}{\sqrt {-e \,x^{2}+d}\, \left (e \,x^{2}+d \right ) \sqrt {e}}\) | \(72\) |
Input:
int((-e*x^2+d)^(5/2)/(-e^2*x^4+d^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
(-ln(x*e^(1/2)+(e*x^2+d)^(1/2))*(e*x^2+d)^(1/2)+2*x*e^(1/2))*(-e^2*x^4+d^2 )^(1/2)/(-e*x^2+d)^(1/2)/(e*x^2+d)/e^(1/2)
Time = 0.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 3.24 \[ \int \frac {\left (d-e x^2\right )^{5/2}}{\left (d^2-e^2 x^4\right )^{3/2}} \, dx=\left [-\frac {4 \, \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {-e x^{2} + d} e x - {\left (e^{2} x^{4} - d^{2}\right )} \sqrt {e} \log \left (\frac {2 \, e^{2} x^{4} - d e x^{2} + 2 \, \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {-e x^{2} + d} \sqrt {e} x - d^{2}}{e x^{2} - d}\right )}{2 \, {\left (e^{3} x^{4} - d^{2} e\right )}}, -\frac {2 \, \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {-e x^{2} + d} e x + {\left (e^{2} x^{4} - d^{2}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e^{2} x^{4} + d^{2}} \sqrt {-e x^{2} + d} \sqrt {-e} x}{e^{2} x^{4} - d^{2}}\right )}{e^{3} x^{4} - d^{2} e}\right ] \] Input:
integrate((-e*x^2+d)^(5/2)/(-e^2*x^4+d^2)^(3/2),x, algorithm="fricas")
Output:
[-1/2*(4*sqrt(-e^2*x^4 + d^2)*sqrt(-e*x^2 + d)*e*x - (e^2*x^4 - d^2)*sqrt( e)*log((2*e^2*x^4 - d*e*x^2 + 2*sqrt(-e^2*x^4 + d^2)*sqrt(-e*x^2 + d)*sqrt (e)*x - d^2)/(e*x^2 - d)))/(e^3*x^4 - d^2*e), -(2*sqrt(-e^2*x^4 + d^2)*sqr t(-e*x^2 + d)*e*x + (e^2*x^4 - d^2)*sqrt(-e)*arctan(sqrt(-e^2*x^4 + d^2)*s qrt(-e*x^2 + d)*sqrt(-e)*x/(e^2*x^4 - d^2)))/(e^3*x^4 - d^2*e)]
\[ \int \frac {\left (d-e x^2\right )^{5/2}}{\left (d^2-e^2 x^4\right )^{3/2}} \, dx=\int \frac {\left (d - e x^{2}\right )^{\frac {5}{2}}}{\left (- \left (- d + e x^{2}\right ) \left (d + e x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((-e*x**2+d)**(5/2)/(-e**2*x**4+d**2)**(3/2),x)
Output:
Integral((d - e*x**2)**(5/2)/(-(-d + e*x**2)*(d + e*x**2))**(3/2), x)
\[ \int \frac {\left (d-e x^2\right )^{5/2}}{\left (d^2-e^2 x^4\right )^{3/2}} \, dx=\int { \frac {{\left (-e x^{2} + d\right )}^{\frac {5}{2}}}{{\left (-e^{2} x^{4} + d^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((-e*x^2+d)^(5/2)/(-e^2*x^4+d^2)^(3/2),x, algorithm="maxima")
Output:
integrate((-e*x^2 + d)^(5/2)/(-e^2*x^4 + d^2)^(3/2), x)
Time = 0.13 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.47 \[ \int \frac {\left (d-e x^2\right )^{5/2}}{\left (d^2-e^2 x^4\right )^{3/2}} \, dx=\frac {2 \, x}{\sqrt {e x^{2} + d}} + \frac {\log \left ({\left | -\sqrt {e} x + \sqrt {e x^{2} + d} \right |}\right )}{\sqrt {e}} \] Input:
integrate((-e*x^2+d)^(5/2)/(-e^2*x^4+d^2)^(3/2),x, algorithm="giac")
Output:
2*x/sqrt(e*x^2 + d) + log(abs(-sqrt(e)*x + sqrt(e*x^2 + d)))/sqrt(e)
Timed out. \[ \int \frac {\left (d-e x^2\right )^{5/2}}{\left (d^2-e^2 x^4\right )^{3/2}} \, dx=\int \frac {{\left (d-e\,x^2\right )}^{5/2}}{{\left (d^2-e^2\,x^4\right )}^{3/2}} \,d x \] Input:
int((d - e*x^2)^(5/2)/(d^2 - e^2*x^4)^(3/2),x)
Output:
int((d - e*x^2)^(5/2)/(d^2 - e^2*x^4)^(3/2), x)
Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.20 \[ \int \frac {\left (d-e x^2\right )^{5/2}}{\left (d^2-e^2 x^4\right )^{3/2}} \, dx=\frac {2 \sqrt {e \,x^{2}+d}\, e x -\sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) d -\sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) e \,x^{2}+2 \sqrt {e}\, d +2 \sqrt {e}\, e \,x^{2}}{e \left (e \,x^{2}+d \right )} \] Input:
int((-e*x^2+d)^(5/2)/(-e^2*x^4+d^2)^(3/2),x)
Output:
(2*sqrt(d + e*x**2)*e*x - sqrt(e)*log((sqrt(d + e*x**2) + sqrt(e)*x)/sqrt( d))*d - sqrt(e)*log((sqrt(d + e*x**2) + sqrt(e)*x)/sqrt(d))*e*x**2 + 2*sqr t(e)*d + 2*sqrt(e)*e*x**2)/(e*(d + e*x**2))